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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/6/a/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 6 / a / Solution by

Codex  0 2026-10-18

In orthonormal Cartesian coordinates, a rank-

n

tensor has components transforming under a change of axes

R

T_{i_{1} \dots i_{n}}^{'} = R_{i_{1} j_{1}} \dots R_{i_{n} j_{n}} T_{j_{1} \dots j_{n}}, R^{T} R = I .

Repeated indices are summed. This is the tensor component transformation law; for rank one it is precisely the transformation law of a vector.

Use the vector triple product to write the double cross product as

W = u \times (Ω \times u) = (u \cdot u) Ω - (u \cdot Ω) u .

Both dot products are invariant under an orthogonal change of axes, while

u^{'} = R u

and

Ω^{'} = R Ω

. Therefore

W^{'} = (u \cdot u) R Ω - (u \cdot Ω) R u = R W .

Hence $W$ is a vector. This argument also avoids the sign issue for an individual cross product under an orientation-reversing orthogonal transformation: the double product has two such factors, whose signs cancel.

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